Introduction to Variational Methods and Finite Elements
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Transcript of Introduction to Variational Methods and Finite Elements
Introduction to variational methods and nite elements
1.2.3. Variational formulations of BVP:
Problem: Sove ax = b x =ba
Reformulate the problem:Consider E = 12ax
2 + bxFind x : E(x) = min
xE(x)
ax b
x
x
1. Rayleigh-Ritz Method:
Consider a dierential equation
Au = u = f(x) (1a)u(0) = u(1) = (1b)
} Functionalan dimensionvector
Consider the functional: E[u] =10
12(u
)2 + fu dx potential energy functional.
Claim: If u : E[u] = minu
E[u] and u satises (1b) then u solves (1).
Proof:
Let u = u + (arbitrary) (0) = 0 = (1) and C2 so we have a 1 parameter family offunctions which are perturbations of u. The function
Nu
E() = E[u + ] =1
0
12
(u
+
)2+ f(u + ) dx
E(0) =1
0
(u+ ) + f dx
=0
=
10
(u) + f dx
= u10
10
(u
f) dx = 0
22
Since is arbitrary we could choose =(u
f)
u f = 0 Euler -Lagrange equation for E[u].
How is this useful? Let us assume that U(x) =N1k=1
kk(x)
basis functions
Then
E() =
10
12
[N1k=1
kk(x)
]2+ f(x)
Nk=1
kk(x) dx
For a min:
0 =E
j=
10
(N1k=1
kk
)j + f(x)j(x) dx
=N1k=1
k
10
kj dx+
10
f(x)j(x) dx, j = 1, . . . , N 1
A = b where Akj =1
0
kj dx bj =
10
f(x)j(x)dx.
Example : u = x2u(0) = 0 = u(1)
}u(x) =
x
12(1 x3)
Assume {j} ={1, x, x2, x3
}
U3(x) = c0 + c1x+ c2x2 + c3x3
U3(0) = c0 = 0 u3(1) = c1 + c2 + c3 = 0U3(x) = ax(1 x) + bx2(1 x)
= a{x x2}1
+ b {x2 x3}2
c2x+ c2x2 = c3x+ c3x3c2x(1 x)c2 = c1 c3
A11 =
10
(1 2x)2 dx =1
0
1 4x+ 4x2 dx = x 2x2 + 43x3
1
0
=13
A22 =
10
(2x 3x2)2dx =1
0
4x2 12x3 + 9x4 dx = 43x3 3x4 + 9
5x5
1
0
=20 45 + 27
15=
215
A12 = A21 =
10
(1 2x)(2x 3x2) dx = 16
b1 = +
10
x2(x x2) dx = 120
23
b2 =
10
x2(x2 x3)dx = 130
13 1616
215
a
b
=
120
130
a = 1
15b =
16
U3(x) =x
30(1 x)(2 + 5x)
Exact:
uex(x) =x
12(1 x3)
Natural and Essential B-C
Notice that the basis functions {i} were required to satisfy the BC u(0) = 0 = u(1). B-C that haveto be forced onto the trial solution are called essential B-C typically these involve the solutionvalues and not the derivatives. In the case of derivative B-C it is possible to build the B-C into theenergy functional to be minimized.
Consider{
u(0) = u(1) = u = f(x)
Old energy functional Boundary Condition
E[u] =
10
12(u)2 + fu dx u(1)
E[u] =
10
uu + fu dx u(1)
= uu10
10
(u f) u dx u(1)
= {u(1) }u(1)1
0
(u f)u dx = 0
u = f and u(1) = but says nothing about u(0)!
Dont have to enforce BC at this endpoint.
Eg. 2:
u = x = f uexact = 2 + 7x2 x3
6u(0) = 2 u(1) = 3
essential natural
Let: U(x) = 2 + x(a1 + a2x) = 2 + a1x+ a2x2
24
satises homogeneous version of essential BC
UN (x) = +N
i=1
aii(x) =N
i=0
aii(x) a0 = 1 0(x) = 2
E(a) =
10
12
(aii
)2+ f
(aii
)dx 3
(aii(1)
)
0 =E
aj=
10
(aii
)j + fjdx 3j(1)
0 =(i,
j
)ai + (f,j) 3j(1)
0 = Aa+ b
A11 =
10
dx = 1 A21 = A12 =
10
2x dx = 1 A22 =
10
4x2dx =43x3
1
0
=43
b1 =10
(x) x dx 3 = 313
b2 =10
(x)x2 dx 3 = 314[
1 11 4/3
] [a1a2
]=[
10/313/4
] a1 = 4312 a2 =
14
U2(x) = 2 +4312
x x2
4
x 0 0.2 0.4 0.6 0.8 1RR 2 2.707 3.393 4.060 4.707 5.333EX 2 2.699 3.389 4.064 4.715 5.333 Built in
2. General calculus of variations:
I[u] =
ba
F (x, y, y) dx y(a) = 0, y(b) = 0
0 = I =
ba
F
yy +
F
y(y) dx
= yF
y
b
a
b
a
[d
dx
(F
y
) F
y
]y dx
ddx
(F
y
) F
y= 0 Euler-Lagrange Eq.
Eg.1 (pu) + qu = fu(a) = u(b) =
I[u] =
ba
p(u)2 + qu2 2fu dx+ u
F = p(u)2 + qu2 2fuF
u= 2qu 2f
F
u= 2pu
x
(F
u
) F
u= 2
{(pu) qu + f} = 0
25
Eg. 2: Sturm-Lioville Eigenvalue problems:
u(a) = 0 = u(b); +(pu) + qu + ru = 0 (1)
Multiply by v and integrateba+(pu)v + quv dx +
baruv dx = 0
= I[u] =
bap(u)2 qu2 dx
baru2 dx
=I1I2
=I1I2 I1I2
I22=
1I2
(I1 I2)
=
ba2puu 2quu ruu dx
baru2 dx
=2
ba
[(pu) + qu + ru
]u dx
baru2 dx
= 0
= 0 the function u0 which minimizes I[u] is an eigenfunction of (1) and = I[u0] is its eigenvalue.
Higher eigenvalues: n = min(u,k)=0
k=1,...,n1I[u]
constrained minimization problem.3. Method of weighted residuals
What do we do for nonlinear or dissipative problems for which potential energy functionals dontexist or cannot be found easily?
Consider a BVP
Lu = f in = (a, b) (1a)u(a) = ; u(b) + u(b) = (1b)
An approximate solution U wont in general satisfy (1a) and we associate with U the so-calledresidual
r(U) = LU fNote that r(u) = 0
exact A whole class of methods are obtained by considering various ways to minimize the residual insome sense, usually:
ba
r(U)i dx = 0
test or weight functions
where U =N
i=1
aii(x) basis function
26
1. Collocation: weight functions = (x xi) basis functions polynomials:
Eg:
u u = 0u(0) = 1, u(1) = 0
(A) Let i(x) =(xi+1 x)which satisfy the homogeneous B-C for the problem
U(x) = 1 x+ a1(x2 x)+ a2 (x3 x)
r(U) = {a12 + a26x} {a1 (2x 1) + a2
(3x2 1)}+ 1
= (3 2x) a1 a2(3x2 6x 1)+ 1
With x1 = 0 and x2 = 1 as collocation points we have(3 11 4
)(a1a2
)=( 11
) U1(x) = 1 x
(311
)(x2 x) 2
11(x3 x)
If we choose x1 and x2 are chosen to be the zeros of the 2nd degree Legendre polynomial
U2(x) = 1 x 14(x2 x)
(16
)(x3 x)
The exact solution is u(x) =(e ex)(e 1) .
x u(x) U1(x) U1 u U2 U2 u U3 U3 u1/4 0.834704 0.837750 3.05103 0.835938 1.23103 0.769965 6.47 1021/2 0.622459 0.636364 1.39 102 0.625000 2.54 103 0.506144 1.16 1013/4 0.349932 0.360795 0.351563 1.63 103 0.269965 8.00 102
(B) With dierent trial functions:
U(x) = 1 x+ a1 sinx+ a2 sin 3xr(U) = ( sinx+ cosx)a1 3(3 sin 3x+ cos 3x)c2U3(x) = 1 x+ 0.017189 sinx+ 0.011045 sin 3x
using zeros of 2nd degree Legendre polys.
Where to collocate?
What basis fucntions to use?
DeBoor, C. and Swartz, B. , Collocation at Gaussian points, SIAM J. Num. Anal. 10 (1973),582606.
27
2. Method of moments: weight functions i = xi.
Eg:
u + u + x = 0u(0) = 0 = u(1)
Let U(x) = a1x(1 x) + a2x2(1 x) + . . .r(U) = x+ a1(2 + x x2) + a2(2 6x+ x2 x3)1
0
r(U) 1 dx = 0 and1
0
r(U) x dx = 0
116 1112
1122
1920
a1
a2
=
12
13
U(x) = x(1 x)(122649
+110649
x
)
uex(x) =sinxsin 1
x
3. Galerkin method: Expand U(x) and w(x) in terms of the same basis functions.
u(x) =N
i=1
aii w(x) =N
i=1
bii
r
(N
i=1
aii
)(N
i=1
bii
)dx = 0
r
(N
i=1
aii
)j dx = 0
Since the bi are arbitrary.
Eg. u + u+ x = 0u(0) = 0 = u(1)
Let U(x) = a1x(1 x) + a2x2(1 x)r(U) = x+ a1(2 + x x2) + a2(2 6x+ x2 x3)1
0
r(u) 1(x)dx =1
0
{x+ a1
(2 + x x2)+ a2 (2 6x+ x2 x3)} x(1 x)x2(1 x) dx = 0
310
320
320
13105
a1
a2
=
112
120
a1 = 7369 a2 = 741
U(x) = x(1 x)(
71369
+741
x
)
ue(x) =sinxsin 1
x
28
Relationship between Rayleigh Ritz and Galerkin:
Rayleigh Ritz u + u = f = x u(0) = 0 = u(1)
I[u] =
10
u2 u2 + 2fu dx
0 = I =
10
2uu 2uu + 2fu dx
=[uu
]10
10
(u + u f) u dx
0 =1
0
(u + u f)u dx
=
10
r(u)u dx
I(a) =
10
(aii
)(aij
)(
aii)(
ajj
+2f
ajj dx
0 =I
ak= 2
{ai(i,
k)
ai(i,k)
+(f,k)}f = x
Rayleigh-Ritz and Galerkin methods are identical for this problem. true in general for linearproblems but not for nonlinear problems.
From WR to the weak form:
WR1
0
(u + u+ x)v dx = 0
u(0) = 0 = u(1) v satisfy homogeneous Dirichlet i.e. v(0) = 0 = v(1)
Say we wanted to express u in terms of functions that are not twice dierentiable then we integrateby parts to throw as many derivatives as needed from u to v.
10
(u + u + x)v dx = uv
1
0
+
10
uv + uv + xv dx = 0.
Now if we let u =
aii and v =
bii we have since v is arbitrary
i
ai(i,j) +
ai(i,j) + (x,j) = 0
Aa = b
where[A]jj = (i,
j) (i,j); b = (x,j)
Identical to Rayleigh-Ritz eqs.
29
What about a natural BC?
(S){
u + u+ x = 0u(0) = u(1) = .
10
(u + u+ x)v dx = 0 v H10